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Lorico [155]
3 years ago
9

Which interval contains a local minimum for the grafted function? [-4, -2.5] [-2, -1] [1, 2] [2.5, 4]

Mathematics
1 answer:
olga_2 [115]3 years ago
7 0

Answer:

Last option [2.5, 4]

Step-by-step explanation:

The local minima of a function are the points where the slope of the curve is zero and the function reaches a minimum value.

If, on the contrary, the function reaches a maximum value, then that point is a local maximum.

Observe, for example, the point (3, -4). Note that a line tangent to that point would be horizontal, that is, of slope m = 0. This point is a minimum of the function, because there are no values x to the right x = 3 or to the left of x = 3 for which f(x)

Now we must search among the options that interval contains at this point.

The intervals [-4, -2.5] [-2, -1] do not contain any local minimum

The intervalor [1, 2] contains a local maximum.

The only interval that contains a local minimum is [2.5, 4], which contains the point (3, -4)

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PLLLLEEEEEEAAASSSSSSSSEEEE
mrs_skeptik [129]

Given:

Line a passes through (2, 10) and (4, 13).

Line b passes through (4, 9) and (6, 12).

Line c passes through (2, 10) and (4, 9).

To find:

Which of the lines, if any are perpendicular.

Solution:

If a line passes through two points, then the slope of line is

m=\dfrac{y_2-y_1}{x_2-x_1}

Line a passes through (2, 10) and (4, 13).  So, slope of this line is

m_a=\dfrac{13-10}{4-2}=\dfrac{3}{2}

Line b passes through (4, 9) and (6, 12).  So, slope of this line is

m_b=\dfrac{12-9}{6-4}=\dfrac{3}{2}

Line c passes through (2, 10) and (4,9).  So, slope of this line is

m_c=\dfrac{9-10}{4-2}=\dfrac{-1}{2}

Product of slopes of to perpendicular lines is -1.

m_a\cdot m_b=\dfrac{3}{2}\times \dfrac{3}{2}=\dfrac{9}{4}\neq -1

m_b\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1

m_a\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1

Therefore, any of these lines are not perpendicular to each other.

8 0
3 years ago
Why do we have to perform a closing procedure in the account cycle?
Finger [1]

Answer:

no

Step-by-step explanation:

7 0
4 years ago
For what value of x is quadrilateral CDEF a parallelogram?
kykrilka [37]

Answer:

x = 4

Step-by-step explanation:

Since the diagonals of a parallelogram bisects each other, therefore:

x + 1 = 2x - 3

Collect like terms

x - 2x = -1 - 3

-x = -4

Divide both sides by -1

x = 4

3 0
3 years ago
Slope of (17,4), (-1,-6)
DENIUS [597]
18/10

M= 9/5

Hope this helps
5 0
3 years ago
Read 2 more answers
Sara wants to find the input value that produces the same output for the functions represented by the tables.
Korolek [52]

Answer:

The input value that produces the same output value in both charts is 2.

Step-by-step explanation:

You are given two functions f(x)=-0.5x+2 and g(x)=2x-3 with tables

\begin{array}{cc}x&f(x)\\-3&3.5\\-2&3\\-1&2.5\\0&2\\1&1.5\\2&1\\3&0.5\end{array}

and

\begin{array}{cc}x&g(x)\\-3&\\-2&\\-1&\\0&\\1&\\2&\\3&\end{array}

First, fill in the second table:

g(-3)=2\cdot (-3)-3=-6-3=-9\\ \\g(-2)=2\cdot (-2)-3=-7\\ \\g(-1)=2\cdot (-1)-3=-5\\ \\g(0)=2\cdot 0-3=-3\\ \\g(1)=2\cdot 1-3=-1\\ \\g(2)=2\cdot 2-3=1\\ \\g(3)=2\cdot 3-3=3

Hence, the second table is

\begin{array}{cc}x&g(x)\\-3&-9\\-2&-7\\-1&-5\\0&-3\\1&-1\\2&1\\3&3\end{array}

The input value that produces the same output value in both charts is 2.

6 0
4 years ago
Read 2 more answers
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